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Lecture Comments (19)

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Post by Swati Sharma on January 29 at 11:34:40 AM

Dear Dr Raffi,

If I am not wrong, this means that if K>1, there are more products that the reactants and if K<1 there are more reactants than the product at EQULIBRIUM. So k would not tell us about the direction of the reaction? K would tell us only if there are more products that the reactants?
SO to find the direction of the reaction we need to first find Q?
So for example if question says if the: Equilibrium constant of the reaction is 25, which direction the reaction would proceed? So I know that there are more products than the reaction, but what about the direction? Does that mean that if k>1 the reaction would go in the forward direction and if K<1 the reaction would go in the reverse direction?
In exam would Q be given? What if Q is not given and they tell us to find the direction of the reaction only with K. So if the Equilibrium constant of the reaction is 25, which way would the reaction proceed? So how  do I find the direction if  Q is not given in the exam? and only with K =25 or may be the question is wrong? I always loose points on these questions in exam I just wondering if K is only to know the concentrations at equilibrium , and Q is fro the direction. But what If Q is not given, only K is given and the question does not ask whether the Products are in greater concentration or the reactants, but instead asks what is the direction of the reaction?
So I thought if K = 25, K>1, there are more products at equilibrium, so the reaction would go towards reactants , since according to Le Chatliers principle  there are more products , so the excess products would be converted to reactants in order to maintain equilibrium. I am really sorry if my question is frustrating , I just want to make sure what I am thinking is right or wrong. I short K will not give us the direction? Q will give us the direction? Would I get a question in exam asking me about the direction of the reaction only with K or q would be provided?


2 answers

Last reply by: Swati Sharma
Mon Jan 29, 2018 10:24 AM

Post by Swati Sharma on January 28 at 11:27:05 PM

Dear Dr Raffi, I am very confused with this question

The equilibrium constant for the reaction Q ? R is 25.
(a) If 50 uM of Q is mixed with 50 uM of R, which way will the reaction proceed: to generate more Q or more R?

The answer says that the reaction would go towards the formation of products but I thought that according to Lechatliers principle when the products are in excess the reaction would go towards reactants to balance each other right?, So why does the answer says that the reaction would go towards products?

3 answers

Last reply by: Professor Hovasapian
Sun Jul 3, 2016 7:27 PM

Post by Kaye Lim on June 13, 2016

At 30:25 where you said as rxn move forward to reach equilibrium, rxn spend its free Energy and delG is risen as reactant is gone.
-->At this point, I dont understand how delG is risen when you take a new measurement of G reactant and G product. I thought for 1 mole of reactant go to product, the process release for example -50kJ of free Energy. So as rxn moves toward equilibrium, x amount of reactant got used, and there are still some reactant unreacted.

Basically, I dont understand how delG risen to 0 as rxn reach equilibrium.
This is my interpretation for delG=0 at equil, please tell me if Im correct. DelG=0 at equilibrium because rate of forward and rate of reverse rxn are equals at equilibrium. Thus, the Energy released for the forward reaction also got used at the same time by the reverse rxn to form reactant. Therefore, overall, change in free E =0 as the Energy released also got used at equilibrium. Is it correct?

1 answer

Last reply by: Professor Hovasapian
Wed Oct 8, 2014 6:04 PM

Post by leala aljehane on October 8, 2014

Hello, Prof.Hovasapian
I just have a question. you said that when   Delta G is positive value that means the product has more energy than the reactants; therefore, we need to input energy to drive the reaction. How can we know that the products have more energy than reactants before we actually start the reaction and get the product? because we just have reactants which group of substances react to form product? I just get confused how can I know the energy for something that I have not yet have it. I hope that I clear my question.

1 answer

Last reply by: Professor Hovasapian
Sat Nov 9, 2013 2:09 AM

Post by Razia Chowdhry on November 8, 2013

Also, where does the concept of ICE come from? And why would the equilibrium concentration be initial conc. + change conc.?

2 answers

Last reply by: Professor Hovasapian
Sat Nov 9, 2013 2:06 AM

Post by Razia Chowdhry on November 8, 2013

For the reaction you did aA+bB-->cC+dD. I don't understand what these small letters stand for? I know the capital stands  for concentration of that species but what about the small letters

1 answer

Last reply by: Professor Hovasapian
Mon May 27, 2013 5:48 PM

Post by marsha prytz on May 27, 2013

so when i put this keq equation into my calculator it does not equal to 20.97. i come up with a completely different number in the 200s. how do i put this calculation into my calculator?

1 answer

Last reply by: Professor Hovasapian
Mon May 27, 2013 5:41 PM

Post by marsha prytz on May 27, 2013

what does the "e" stand for in = e^3.043?

Thermodynamics, Free Energy & Equilibrium

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Thermodynamics, Free Energy and Equilibrium 1:03
    • Reaction: Glucose + Pi → Glucose 6-Phosphate
    • Thermodynamics & Spontaneous Processes
    • In Going From Reactants → Product, a Reaction Wants to Release Heat
    • A Reaction Wants to Become More Disordered
    • ∆H < 0
    • ∆H > 0
    • ∆S > 0
    • ∆S <0
    • ∆G = ∆H - T∆S at Constant Pressure
    • Gibbs Free Energy
    • ∆G < 0
    • ∆G > 0
    • Reference Frame For Thermodynamics Measurements
    • More On BioChemistry Standard
    • Spontaneity
    • Keq
    • Example: Glucose + Pi → Glucose 6-Phosphate
  • Example Problem 1 40:25
    • Question
    • Solution

Transcription: Thermodynamics, Free Energy & Equilibrium

Hello and welcome to Educator.com and welcome back to biochemistry.0000

Today, we are going to start our discussion of a profoundly important topic- bioenergetics.0004

The energy transfers that take place in biological systems.0013

Today, we are going to be discussing the thermodynamics, free energy and equilibrium.0017

I cannot overestimate how important this topic is.0021

In a biochemistry course, you are going to see a lot of information.0026

It is a ton of information, as you already know, and there is certainly a lot more to come.0031

If you do not walk away with anything else from this course, an understanding of bioenergetics will serve you for many years to come, particularly those of you that go on to graduate school and are looking for research careers.0037

Understanding what is happening energetically will allow you to understand pretty much everything that is happening.0050

In any case, let’s just go ahead and jump right on in and get started, and see what we can do with this magnificent, magnificent, beautiful topic.0058

OK, another comment that I want to make really quickly, the problems, as far as this unit is concerned, this bioenergetics, there are going to be several lessons.0067

I will be doing some occasional examples and problems during the lessons, of course, but I am going to save the bulk of the problems that I actually do to the end.0076

In a series of, maybe 1, 2, 3 lessons, I would like to do them all at once, one after the other, one after the other.0085

If there is something that is not exactly clear, do not worry.0091

When we actually do the problems, it is a great way to consolidate; and in this particular case, I thought it would be great to do a whole bunch of them at the end.0095

That way, we deal with concepts; and it gives us a chance to actually review as we do the problems.0101

With that, let’s get started; let’s start with a reaction.0108

You know what, I think I will do blue; blue is always nice.0113

Let’s start with a reaction, and the glucose - I am sorry - the reaction we are going to start with is going to be glucose plus an inorganic phosphate goes to glucose-6-phosphate.0119

OK, let’s go ahead and draw this out real quickly.0140

We have got our glucose molecule; that is that.0144

Let’s go ahead and make it beta here; so, this is down.0149

This is up; this is down, and, of course, we have our CH2OH.0154

Now, plus inorganic phosphate, phosphate, so POOH, I will go ahead and put the H on there; and it goes to glucose 6-phosphate.0159

It looks like this; that is that.0173

That is that; that is here, glucose-6-phosphate.0178

This is the 6 carbon, and, of course, it is going to look like that.0184

That is a structural representation of the reaction.0192

Now, this happens to be the first step of glycolysis; well, you know what, I do not need to write that part down.0195

This happens to be the first step of glycolysis, and this is a reaction that we definitely want to move forward - very, very, important - that this reaction move forward and move forward easily.0201

OK, here is the interesting thing; we can write down any reaction we want.0212

We can put anything on the reactant side, and if it is reasonable or seems to make sense, we can put anything on the product side; but just because we write down a reaction - excuse me - just because we write an equation, it does not mean that it happens.0218

Just because we write it down, it does not mean that it is actually going to happen.0249

OK, in discussing thermodynamics, we speak of spontaneous processes, or more specifically, spontaneous reactions.0253

OK, now, spontaneous does not have the same meaning as in normal daily speech.0287

Spontaneous means it wants to happen; that is the important thing.0297

It wants to happen naturally, and if circumstances are right, it does happen naturally and easily.0305

It does happen naturally, in other words, without external pressure.0331

So, when we talk about a spontaneous process, we are making a thermodynamic statement.0346

We are saying that this reaction, all else being equal, it wants to happen; and if the reaction actually starts, and the circumstances are good for it to happen, it will happen without us doing anything about it.0350

A spontaneous does not refer to speed; it does not talk about how fast this reaction is going to take place.0362

That falls under the domain of kinetics; thermodynamics just speaks about the potential for something happening, the extent to which it wants to happen.0367

Kinetics is the domain that discusses whether we can make it happen and how fast we can make it happen; so those are two different things.0376

OK, now, our thermo is a study of the extent to which a reaction wants to happen and can happen naturally; that is all thermodynamics is.0385

Alright, now, in any reaction, nature - maybe I should capitalize nature and be respectful - in any reaction, nature wants 2 things.0400

All of our observations, all of the things that we have done, all of the things that we have observed, all the data we have put together, has led us to a couple of conclusions.0421

One of those conclusions regarding natural processes is the following regarding reactions.0429

In going from reactants to products, a reaction wants to release heat.0435

It wants to release heat; in other words, the bonds in a molecule represent a certain amount of energy.0459

Ideally, if it can move, if the reactants can move and become a product where the bonds are more tightly held together and in that process, more stable, its energy is reduced.0471

So, it actually wants to release heat naturally; that is what it would like to do under normal, ideal circumstances.0485

Now, ΔH of the reaction, the enthalpy of reaction, that is the quantitative measure of this tendency.0494

OK, ΔH reaction is the quantitative measure of this tendency at constant pressure.0511

At constant pressure, the heat that reaction releases or gains is called the enthalpy- that is all it is.0530

It is a quantitative measure of the release or the gain of heat in a reaction, but under ideal conditions, it wants to release heat.0542

Now, 2- the reaction wants to become more disordered; things fall apart.0551

Things do not just naturally go from a state of disorder to order without some external help, but things will naturally go from a state of order to disorder, naturally, without us doing anything about it.0570

That is the nature of the universe that is the second law of thermodynamics.0582

That entropy always increases in a natural process, has to.0586

So, under ideal conditions, a reaction would like to become more disordered as it moves from reactant to product.0591

That favors the forward movement of the reaction, wants to become more disordered.0599

OK, ΔS entropy - I will just put rxn of the reaction - is the quantitative measure of this tendency.0605

OK, now, when we have a ΔH, which is less than 0, which is negative, the enthalpy of the products minus the enthalpy of the reactants, when it ends up being negative, that means it has given up heat.0630

This is 'releases heat"; these are ideal conditions for a reaction to easily move forward.0645

OK, so I will just call it that; these are ideal conditions.0655

When the ΔH is greater than 0, it requires heat, or it absorbs heat from its surroundings, absorbs energy- these are less than ideal conditions.0658

Notice, I am calling something ideal, but I am not excluding the other.0670

When I say that the ΔH is greater than 0, I am not saying that it does not happen.0675

I am saying that it is less than ideal conditions, so we will leave it at that for the moment.0679

Now, ΔS, when ΔS of a reaction is greater than 0, OK, when disorder increases as you go from product, those are ideal conditions.0684

In other words, the entropy, the measure of disorder of the products is greater than the measure of the disorder of the reactants.0700

Product minus reactants, you get a positive number.0706

Those are ideal conditions for a reaction to move forward naturally, without any help, and ΔS less than 0, well the disorder decreases.0710

Those are less than ideal conditions; those are less than ideal.0725

OK, now, let’s go ahead and go to red.0732

These 2 properties, the enthalpy and the entropy, these 2 properties, ΔH and ΔS plus the third property, which is temperature, they contribute, they combine and they compete in such a way as to give rise to a single - I am not going to use that one - to give rise to a single number that tells us whether a reaction wants to move forward naturally, whether a reaction wants to and can move forward naturally.0737

OK, property of ΔH and ΔS plus this third property, temperature, they contribute, they combine and they do compete in such a way as to give rise to a single number that tells us whether a reaction wants to and can move forward naturally.0842

That number is the ΔG, and I will write the equation first.0856

ΔG is equal to ΔH - T, Δ-S- profoundly, profoundly, profoundly important relation in thermodynamics.0864

If I have a reaction, I can measure its enthalpy, the change in the enthalpy.0873

I can measure the change in entropy, and at a given temperature, these 3 things, the enthalpy, the temperature, they combine; and we call this combination the Gibbs free energy.0878

It is a single number that tells us whether a reaction will move forward spontaneously under the ideal conditions, under the right circumstances, or whether it cannot; and if it cannot, that means that we have to do something to make it happen- that is it.0888

ΔG is called the Gibbs free energy.0901

Oh, I should say at constant pressure - that is important - and under physiological conditions, the pressure is going to be constant.0913

So ΔG is called the Gibbs free energy, and it tells us if a reaction as written - and when we say as written, we mean left to right, reactants to products - as written is thermodynamically favorable.0926

It does not tell us that it is going to happen; it tells us that it is favorable, that all else being equal, it will happen- that is all that it says.0956

It does not say about how fast or if it actually does happen.0964

If I put hydrogen and oxygen into a tank and I just let it sit there, it is not going to do anything.0968

The hydrogen and oxygen are not going to combine to form water.0975

The thermodynamic says they really, really, really want to form water very badly, but it is not going to do just because I want it to be, just because it is thermodynamically favorable.0978

There is something else I have to do to make it happen; I have to ignite it with a spark.0987

Put a little bit of energy in to it just to get it over that hump, that activation energy, then everything will be fine.0992

The reaction will sustain itself, and it will go very, very far forward and release a hell of a lot of energy; but it does not mean it is going to happen.0998

So, this thermodynamic just speaks about its tendency to happen.1006

OK, now, ΔG less than 0, so a negative ΔG, that is favorable.1010

OK, that is favorable; it is natural.1018

That is called spontaneous; now, if the Δ G happens to be greater than 0, it is not favourable.1023

It is not, well, yes, it is not favorable, what we say, that it is spontaneous in reverse.1038

You remember when you switch directions, if you go from left to right and like ΔG or ΔH are just positive, if you go from right to left, you just switch the sign; it is negative.1045

If ΔG is greater than 0 for a reaction as written on the page, well, that means it is negative in the other direction.1054

That means it is spontaneous in other direction, so spontaneous in reverse direction.1060

OK, let’s see; now, OK, whenever we do anything in science, we need a reference frame for all comparative measurements.1071

If we are going to make a bunch of measurements of a bunch of different reactions, we need conditions.1101

We need to set up conditions, so that we run these reactions under the same conditions, so that we can actually compare one reaction to another or one circumstance to another.1107

We cannot just have the temperature be one thing in one experiment and then the temperature be another thing in another experiment, the pressure be different, concentrations be different.1116

In science, we need a reference frame for all comparative measurements; this is profoundly important.1123

OK, in thermodynamics, OK these properties are calculated according to a standard, according to the following standards.1131

When we run these reactions to measure these quantities, this is the standard that we follow.1159

OK, 25°C or 298K, that is the temperature standard.1166

All aqueous reagents - I will not say reagents, I will say all aqueous species - are at 1M concentration.1174

All gaseous species are at 1 atmosphere pressure, and the super script that we use is this little circle there like a little degree sign on top of the thermodynamic quantities; and that lets us know that these numbers are under standard conditions.1194

Under standard conditions of temperature and pressure, our equation becomes ΔG 0 = Δ H 0 - or whatever you want to call that thing, it is up to you - -T ΔS 0.1220

That just means that we are doing these things under standard conditions.1251

When you look in a chemistry book of the tables, on the back, the thermodynamic tables, you are going to see these.1256

Those numbers, they were all done under standard conditions because we need a common frame of reference, so that we can make some intelligent choices about what it that is going on- that is it.1260

Now, let’s talk about some units, so ΔG is in kJ/mol.1272

ΔH is also kJ/mol; usually, we will not write out the Joule, we just say kJ.1288

ΔS is J/mol-K, so it is very, very important to watch your units.1298

Remember, if you are using this equation, ΔH is going to be in kJ/mol; that is going to be the number most often reported.1307

That is the unit; ΔS is going to be J/mol-K.1315

So, either you covert this to kilojoules or you convert this to Joules, whichever is more convenient.1318

I tend to just convert the ΔH to Joules; it just depends on the unit that you are interested in.1323

You will get an answer in Joules, the delta for the ΔG, and then you can just go back to kilojoules if you want to; it is up to you.1328

Remember to convert- very, very important in all of these things.1335

Watch your units; remember to convert kJ to J or J to kJ, as necessary.1339

OK, now, a little bit more about standards.1354

Now, many biochemical reactions, they involve hydrogen ion, involve H+.1359

That is an aqueous species, and we said that aqueous species, under standard conditions are 1M.1372

Now, well, 1M H+ means that the pH is 1.1378

When running a particular reaction under standard conditions, the hydrogen ion concentration for a reaction that involves hydrogen ion, which is the majority of biochemical reactions, the pH is going to be 1.1395

However, in biochemistry or physiological conditions, the pH is not 1.1405

The pH hovers around 7, 7 to 7.3, 7.4- something like that1409

There is a transformed standard - a biochemical standard, if you will, not a chemical standard - where what we do is we set the concentration of the hydrogen ion to pH7.1414

The concentration of - we set the pH to 7 - the concentration of hydrogen ion is 10-7M, so it is a slightly modified standard.1424

Let me see, so many biochemical reactions involve H+, and 1M H+ means pH1, but physio pH = 7.1434

The transformed biochem standard calls for a hydrogen ion concentration equal to 10-7.1452

So we set the pH to 7, as well as, and since many reactions, we are going to be talking about bioenergetics, which means we are going to be talking about adenosine triphosphate.1480

Adenosine triphosphate tends to require magnesium.1492

In addition to the hydrogen ion concentration being 10-7 as well as the magnesium ion concentration, we usually set to 1mM.1497

This is called the biochemical transformed standard, whatever it is you want to call it; and you will see it written like this.1510

Instead of ΔG 0, we will go ahead and write it as ΔG 0 that.1519

When we do our numbers, the numbers that you see in biochem texts, they are under the transformed biochem standard.1525

Those are the numbers that we are going to be using in our problems.1531

OK, now that we have talked about standards a little bit, let's go ahead and discuss what it is that we actually mean by spontaneous.1534

Let me do black, now; just really briefly, let's discuss spontaneity, and we are going to use an energy diagram- our favorite thing.1543

We said ΔG less than 0, it is spontaneous as written.1560

We have ΔG greater than 0; it is spontaneous in the reverse direction.1571

OK, here is what this means; I have an energy diagram like this, energy on the Y axis.1580

This is the reaction coordinate; let's go ahead and put reactants there, products there- something like that.1589

This difference right here, this is going to be the ΔG.1597

OK, in this particular case, ΔG, which is G final minus G initial, that is going to be less than 0.1600

This, right here, from this point to this point, this is the activation energy; that is the domain of kinetics.1608

OK, that is it, so this is the energy diagram for a spontaneous reaction.1617

It releases energy and exactly what you think, exactly what you have seen before.1621

I am sure you have seen it 100 times; we will make it 101.1628

Excuse me, and something like that, let's go here.1634

Our ΔG, this is going to be a +ΔG because the energy content of the products is higher than the energy content of the reactants.1640

And again, you have your activation energy- that is it.1648

This is the energy diagram for spontaneous and non-spontaneous or spontaneous in reverse.1652

Now, again, we had mentioned that the term spontaneous refers to thermodynamic potential and tendency.1660

It says nothing about how fast it does or if it actually does so.1671

So, let me go ahead and just draw one more little energy diagram.1675

Now, I do not want it to be even; let's go ahead and make it exergonic.1681

Exothermic, endothermic, heat in, heat out- in the case of ΔG, when we are talking about ΔG, it is exergonic, endergonic.1688

OK, now, these 2 places right here, energy of the reactants and energy of the products, that is the domain of thermodynamics.1697

This right here, the activation energy and the particular path that the reaction takes in order to go from reactants to products, that is the domain of kinetics, not the same thing.1712

Again, when we talk about ΔG, we are talking about the tendency to go.1721

We are not talking about the fact that it is actually going to go and how it is going to go or how fast; that is kinetics.1726

OK, alright, now, let's go ahead and go back to blue.1733

Now, say a reaction, a particular reaction has a ΔG that is less than 0 - I will go ahead and use that little standard - and starts to move forward.1742

So, let's say we have a reaction, ΔG is favorable, and it starts to move forward.1753

OK, we know from general chemistry that as a reaction starts to move forward, eventually, the reaction will reach equilibrium, right?1763

All systems tend toward equilibrium, or the forward and the reverse reaction are the same rate.1773

OK, now, as the reaction moves forward - excuse me - as the reaction moves forward, it spends this free energy.1779

Remember, free energy is a measure of the tendency to move forward.1796

It is telling me that the reaction wants to be on the right, not on the left, a -ΔG.1799

Well, when it starts to actually move forward, well, now, it is spending that free energy.1805

Now, at any given moment, as the reaction is moving forward, if you are to measure the G of the products and the G of the reactants, the ΔG, again, the ΔG will have risen.1809

Now, you are losing free energy; that is what you are doing as you are tending toward equilibrium.1819

As the reaction moves forward, it spends this free energy as it moves toward equilibrium.1824

ΔG is also a measure of the extent to which a reaction as written is not yet at equilibrium.1842

So, the lower your ΔG, let’s say you have a ΔG of -10, let’s say you have a ΔG of -50, that reaction that has a ΔG of -50, that means it is further away from equilibrium.1878

It has more energy to spend as it makes its way toward equilibrium.1890

Once it reaches equilibrium, you are done; the reaction has done all the work that it is going to do for you.1894

It is giving you all the energy that is going to give you; now, the ΔG is 0.1899

We will talk more about that later, of course; now, let’s go to red.1904

We also know another quantitative measure of equilibrium, quantitative measure related to equilibrium, I should say.1912

We have the Keq, the equilibrium constant; it is another quantitative measure of where a reaction wants to be at equilibrium.1938

That is what the number tells you; is the reaction really far forward to the right at equilibrium?1946

Is it mostly product, virtually no reactant?1951

Is it to the left, a lot of reactant, no product; or is it somewhere in between?1954

That is what the Keq does; that is what the ΔG does.1959

OK, just a quick recap; for the reaction AA + BB goes to CC + DD, we know that the Keq is equal to - well, reactant, law of mass action - so, AA the constant - oops - those are the reactants.1963

It is going to be products over the reactants; it is going to be CC, DD, over Aa, and it is going to be Bb.1988

So, that is the general expression for Keq.1999

These are the concentrations at equilibrium- very, very important - at equilibrium.2002

OK, now, well, ΔG is a measure of the extent it is from equilibrium.2007

The Keq is a measure of the extent to which a reaction is far from equilibrium.2014

They happen to be 2 different quantitative measures, but they are measuring the same thing.2018

There must be a relation between them- there is.2021

Here, and here it is; let me go back to black here.2024

The ΔG is equal to -RT ln Keq, where R is equal to 8.315J/mol-K, and T is equal to the temperature, the absolute temperature, which means the temperature in Kelvin- that is it; that is all it is.2030

Now, let’s return to our reaction; let’s return to our initial reaction.2055

We said we had glucose plus inorganic phosphate forming glucose 6-phosphate.2068

Well, the ΔG standard for this happens to be +13.8kJ/mol.2077

This is endergonic; this is not spontaneous as written.2087

It is spontaneous in reverse.2090

What this tells me is that glucose, that under standard conditions, 1M concentration of all these things, this reaction is not going to want to move forward naturally; it is going to want to move backward naturally.2093

The glucose 6-phophate is going to want to break up into inorganic phosphate and glucose, so this is not spontaneous as written- excuse me.2103

In other words, under standard conditions, this reaction will not go forward; we want it to go forward.2124

This is not a thermodynamically favorable reaction.2149

Let’s see what we have got; let’s go ahead and do some calculations here.2155

Once again, we have got glucose plus inorganic phosphate; the equilibrium is actually going to look like this.2159

We know we write arrows of different lengths to show us that their equilibrium here, once it comes to equilibrium, is actually going to be toward the left.2167

It is going to be a very little product; it is going to be a whole bunch of reactant.2175

Again, under standard condition, when we start with 1M concentration of each of these species, again, we need a standard to work with.2180

Well, when we go ahead and put this 13.8kJ/mol into our equation, what we end up is the following.2188

We can rearrange this equation to solve for the Keq- excuse me.2197

It is going to be E to the minus ΔG / RT = E-13,800J - I have to convert to Joules because R is in J/molK - over 8.315 x 298K, what you end up with is 3.8 x 10-3.2203

Under standard conditions, it is confirmed; this Keq is very, very small.2234

That means it is going to be to the left; that is the whole idea.2239

Small Keq means not a lot of product, a whole bunch of reactant.2246

This is just another numerical measure, and these are related by R and T; that is all that is happening here.2251

You just have 2 quantitative measurements to let you know where equilibrium lies for a reaction as written.2257

OK, let’s see; let’s go ahead and do another example here.2264

This time, we will go ahead and do the hydrolysis of adenosine triphosphate.2270

H2O goes to adenosine diphosphate plus inorganic phosphate.2275

In this particular case, the ΔG under standard conditions is -30.5kJ/mol- highly exergonic, highly thermodynamically favorable.2280

This reaction wants to take place; OK, it wants to move forward.2293

Now, this means that you have 30,500J of free energy that you can use to do some kind of work, whatever that biological work is, whether it is transporting some solute across the membrane, whether it is contracting a muscle, whether it is building a nucleic acid or building a carbohydrate polymer- whatever it is.2299

Per mole of adenosine triphosphate, I have 30500J of energy that I can use.2322

Now, when I measure this Keq, you know what, I am just going to - that is OK, that is fine - I was going to not write the eq part, but that is OK.2329

When I go ahead and put it into that equation, it is going to be E to the negative.2338

Now, the equation is -ΔG / RT, so -0ΔG is negative.2345

I have to include both of both negatives, so 30,500 all over the 8.315, 298K; and what you end up with is a very huge number 2.2 x 105.2353

The fact that the Keq is huge, that confirms the fact that there is a whole bunch of product, the numerator, very little reactant and denominator- that is all that is going on here.2374

At equilibrium, it is going to look like this: ATP + H2O, that way, that way, ADP + PI.2385

Excuse me; that is what the equilibrium is going to look like.2398

OK, let’s go ahead and finish off by doing a little bit of an example.2403

I will consolidate everything that we talked about - equilibrium, free energy - with a typical example that we are going to see, actually, several times when we get to the problem section towards the end of this particular bioenergetics unit.2408

Let’s see what we have got.2424

OK, this says when glucose 1-phosphate is incubated at 25°C with the enzyme phosphoglucomutase, the following reaction takes place.2426

Glucose 1-phosphate is converted to glucose 6-phosphate.2435

OK, the ΔG for this reaction is -7.54kJ/mol, so it is spontaneous as written.2439

The question they are asking is, if the initial concentration of the glucose 1-phosphate is 0.15M, calculate the concentrations of each species when the reaction has come to equilibrium.2446

OK, we are given our ΔG; we are given our initial concentration of glucose 1-phosphate.2457

We want to find the equilibrium concentrations; OK, let’s see what we can do here.2467

This is essentially an equilibrium problem, and as such, we are going to end up making something called an ice chart, if you remember from general chemistry.2473

The plan for this particular problem - let me do it in blue - the plan is we have the ΔG; they give us the ΔG.2482

From the ΔG, we are going to find the Keq, and then after that, we are going to do an ice chart; and ice just means initial change equilibrium, one of those charts that we used to do in general chemistry to track where things, how things start, where they end up; and then do the calculation based on the Keq.2491

Let’s go ahead and do that; we start with our equation, which is, of course, ΔG.2514

I am not going to go ahead and keep writing these super scripts over and over and over and over again.2521

You know that we do all of these calculations under the transformed biochemical standard, so I am just going to write ΔG = - RT ln Keq.2525

When I rearrange Keq = E to the -ΔG / RT, when I go ahead and put these values in, Keq = E to the minus.2538

Let's go ahead and do, we said that the ΔG was -7.54 kJ/mol, so -7540, that is kJ/mol / RT, which is 8.315, and we have 298 because we are running this under standard conditions 25°C, OK, what you end up with is E3.043, and you get 20.97.2552

Our Keq is 20.97; now that we know what the Keq is, now, we will do the equilibrium part.2586

OK, let’s go ahead and write our reaction; glucose 1-phosphate is going to reach some sort of an equilibrium with glucose 6-phosphate.2593

We are going to have an initial concentration; we are going to have the change that takes place, and then we are going to have an equilibrium concentration.2605

This is what we are concerned with ultimately.2611

Well they said we started off with 0.1M glucose 1-phosphate, and we have none of this, so that is our initial concentration2614

Well, we know that the reaction is going to move forward; there is none of this.2622

So, it is going to move forward; that means glucose 1-phosphate is going to deplete.2625

It is going to deplete by a certain amount x; well, the amount of depletion - this is a 1 to 1 ratio - is going to be exactly the amount of glucose 6-phosphate that shows up.2629

This is going to be +x; that is the change that takes place.2638

The equilibrium is the initial plus the change that took place, so 0.15 - x and x.2642

These are the values that I need, and let me go ahead and do the calculation on the next page.2651

I have Keq = x / 0.15 - x, products over reactants at equilibrium.2660

Well, this happens to equal 20.97; I know what the equilibrium constant is.2673

I go ahead and solve this equation x = 20.97 x 0.15 - x.2679

I will go ahead and put 3 dots; you guys go ahead and do the algebra.2687

What you end up with is x = 0.1432M.2692

That is going to equal the equilibrium concentration of the glucose 6-phosphate.2699

Now, x, I am sorry; the equilibrium concentration of the glucose 1-phosphate, if you go back, it said it is a 0.15 - x.2705

When we do that, we get 0.0068M equals the concentration of the glucose 1-phosphate- there you go; that is our final answer.2716

We started with the ΔG; we convert it to Keq.2729

We just did an equilibrium problem, and this confirms the fact that most of the glucose 1-phosphate has been converted to glucose 6-phosphate.2732

There is virtually no glucose-1-phosphate left over.2740

OK, that is it.2744

Thank you for joining us here at Educator.com.2748

We will see you next time, bye-bye.2750